The main anion `O^(-)` is isoelectronic with

Options

A. `N^(2-)`
B. `F^(-)`
C. `N^(3-)`
D. `Ne`

Answer

To determine which species is isoelectronic with the anion \( O^- \), we need to follow these steps:

### Step-by-Step Solution:

1. **Identify the Electron Configuration of \( O^- \)**:
- Oxygen (O) has an atomic number of 8, which means it has 8 protons and, in its neutral state, 8 electrons.
- The anion \( O^- \) has gained one additional electron, resulting in a total of 9 electrons.

2. **Understand the Concept of Isoelectronic Species**:
- Isoelectronic species are atoms or ions that have the same number of electrons. In this case, we are looking for a species that also has 9 electrons.

3. **Evaluate Each Option**:
- **Option A: \( N^{2-} \)**:
- Nitrogen (N) has an atomic number of 7, meaning it has 7 protons and 7 electrons in its neutral state.
- The \( N^{2-} \) ion has gained 2 electrons, resulting in a total of 9 electrons (7 + 2 = 9).
- **Option B: \( F^- \)**:
- Fluorine (F) has an atomic number of 9, which means it has 9 protons and 9 electrons in its neutral state.
- The \( F^- \) ion has gained 1 electron, resulting in 10 electrons (9 + 1 = 10).
- **Option C: \( N^{3-} \)**:
- Again, nitrogen has 7 protons and 7 electrons in its neutral state.
- The \( N^{3-} \) ion has gained 3 electrons, resulting in a total of 10 electrons (7 + 3 = 10).
- **Option D: Ni (Nickel)**:
- Nickel (Ni) has an atomic number of 28, which means it has 28 protons and 28 electrons in its neutral state. This is far more than 9 electrons.

4. **Conclusion**:
- The only species that has the same number of electrons (9) as \( O^- \) is \( N^{2-} \).
- Therefore, the answer is \( N^{2-} \).

### Final Answer:
The main anion \( O^- \) is isoelectronic with \( N^{2-} \).

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